PLEASE HELP ASAP!!!

You want to lift a heavy box with a mass L = 56.0 kg using the two-ideal pulley system as shown. With what minimum force do you

Olin [163]

The minimum force required to lift the box at constant velocity is determined as 274.4 N.

<h3>Minimum force required</h3>

The minimum force required to lift the box at constant velocity is the tension in one of the pulleys, and the magnitude is calculated as follows;

2T = mg

where;

m is mass of the box
T is the minimum force required

2T = mg

T = mg/2

T = (56 x 9.8)/2

T = 274.4 N

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3 0

2 years ago

Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth

sashaice [31]

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, $a=1\ AU=1.496\times 10^{11}\ m$

Mass of the Earth, $m_e=5.972\times 10^{24}\ kg$

Mass of the planet, $m_p=2m_e=11.944\times 10^{24}\ kg$

Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.

$T^2=\dfrac{4\pi^2}{Gm_p}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3$

$T=1.28\times 10^{10}\ s$

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.

4 0

3 years ago

A lightening bolt may carry a current of 11800 A for a short period of time. The permeability of free space is 1.25664 × 10−6 T

Deffense [45]

Answer:

The resulting magnetic field is 5.021 x 10⁻⁵ T

Explanation:

Given;

current in the lightening bolt, I = 11800 A

distance from the bolt, r = 47 m

permeability of free space, μ₀ = 1.25664 × 10⁻⁶ T· m/A

Assume lightening bolt as long straight conductor, then the resulting magnetic field will be calculated as follows;

$B = \frac{\mu_oI}{2 \pi r}$

where;

B is the resulting magnetic field

I is the current in the bolt

r is the distance from the bolt

Substitute the given values and calculate B

$B = \frac{\mu_oI}{2 \pi r} \\\\B = \frac{1.25664 *10^{-6}*11800}{2 \pi (47)} \\\\B = 5.021 *10^{-5} \ T$

Thus, the resulting magnetic field is 5.021 x 10⁻⁵ T

4 0

3 years ago

1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very di

Mama L [17]

Answer:

Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :

Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.

Cat skeleton: cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.

6 0

3 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 4.0 × 106 m/s moves through a uniform magnetic fi

Natasha2012 [34]

Explanation:

It is given that,

The horizontal and vertical component of velocity of an electron is:

$v=(2.6i+4j)\times 10^6\ m/s$